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In mathematics one defines:

$\left(\begin{array}{c}n\\k\end{array}\right)=\displaystyle\frac{n!}{k!\cdot (n-k)!}$

This is the number of combinations of $k$ elements from a collection of $n$ elements.

I was wondering if it is possible to prove that the result is an integer in a formal way. It's easy to see that $k!$ in the denominator divides the numerator, but how do you prove that the denominator $k!$ also divides the product of all elements between $n-k+1$ and $n$?

hardmath
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willeM_ Van Onsem
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  • Basically what you want to know is why $$ k!: | : n(n-1)(n-2)\cdots(n-k+1) $$ right? One approach is looking at prime factors. Show that for any prime, among all $k$ long products of consecutive (natural) numbers, $k!$ is (tied for) the one with least divisibility by $p$. – Arthur Oct 07 '13 at 13:42
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    Hint: Apply complete mathetical induction to prove $${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k} $$ – gammatester Oct 07 '13 at 13:44
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    You say that $\binom{n}{k}$ is the number of combinations of $k$ elements from a collection of $n$ elements [1]. If $x$ is equal to an integer, is it not a proof that $x$ itself is an integer? A formal way to express [1] is Newton's binomial theorem. – Siméon Oct 07 '13 at 13:47
  • Excuse my naivety, but for a list of consecutive numbers of length N, won't there alway be a number in the list divisible by each of $1 \dots N$ ? – stevemarvell Oct 07 '13 at 13:50
  • @stevemarvell yes, but the same number might be used in two different cases, so you should rule out this case. – mau Oct 07 '13 at 13:55
  • @mau we clearly know that it works, so there must be some more "English" which I could have used to expand the explanation. Clearly you mean something like having a number like 4, which could have candidate factors 4 and 2. If this is the case, then the length of the list of number must be at least 4, in which case there must be two candidates for 2. – stevemarvell Oct 07 '13 at 14:01
  • See this question. – Will Orrick Oct 07 '13 at 14:12
  • @stevemarvell in the sense that it's true that in {28, 29, 30, 31, 32, 33} there is a multiple of 5 and a multiple of 6, but they are the same number. So you cannot just make a simple association, but at least you should work with the factors. – mau Oct 08 '13 at 12:10
  • @mau I can see the point, but in this case it doesn't matter. I know that's not a very formal thing to say. In fact, in any list with a prime number in it, there will exist a number for which the denominator is applied at least twice. It strikes me that you example could only have been contrived (for want of a better word) if the list contained a number which had factors of both. The one I've been toying with is ${13,14,15,16,17}$. I can't generate an example where "cancelling" twice or more times with the same number causes a problem. But then that's to be expected. – stevemarvell Oct 08 '13 at 12:20
  • @stevemarvell it's clear that eventually the result must be an integer, since there's a combinatorical way to prove it. I only wanted to say that the naive argument has to be fleshed a bit. – mau Oct 08 '13 at 13:19
  • @mau I agree. It's the fleshing that I can't prove. Thanks for the indulgence. – stevemarvell Oct 08 '13 at 13:29
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    @stevemarvell when I stumbled into this problem last year, and searched for a non-combinatorial solution, I eventually used the counting argument for numbers of the form $p^m$ (you need also power of primes: for example for $n=13$ you first show that there is one multiple of 8, then that there are a total of three multiples of 4, and last that there are a total of six multiples of 2) – mau Oct 09 '13 at 07:59

1 Answers1

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$$ \frac{n!}{k!\cdot (n-k)!} =\frac{n(n-1)\cdots(n-k+1)}{k!} =\frac{(n-k+k)(n-1)\cdots(n-k+1)}{k!}\\ =\frac{(n-1)\cdots(n-k)}{k!}+ \frac{(n-1)\cdots(n-k+1)}{(k-1)!} $$ Now use (double) induction.

This is the relation that defines Pascal's triangle: $${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}.$$

lhf
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